题目如图点PPP为⊙O\odot O⊙O外一点,过点PPP作⊙O\odot O⊙O的两条切线,切点分别为AAA、BBB,过点AAA作PBPBPB的平行线,交⊙O\odot O⊙O于点CCC,连接PCPCPC交⊙O\odot O⊙O于EEE,连接AEAEAE并延长交PBPBPB于KKK,求证:PE⋅AC=CE⋅KBPE\cdot AC=CE\cdot KBPE⋅AC=CE⋅KB.知识点:切割线定理章节:图形的性质 / 圆